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default Problem Set 5.3 | Universal Gravitation Jonathan Corbett 25 March 2025

Problem Set | Universal Gravitation

1. A 70.0 kg human standing on Earth

Calculating the force of attraction between the human and Earth's center of gravity requires G, (a) Earth's radius, and the masses of the two bodies. As the human is standing on the surface of the Earth, the distance between their centers of mass is simply Earth's radius (ignoring the height of the human)

|3 sf|$F_g = G \frac{m \dot M}{r^2}$| |-:|:-| |$m = 70.0 kg$
$M = 5.970x10^{24} kg$
$r = 6\,378x10^3 m$|$F_g = 6.673x10^{11} \left($ \frac{(70.0 kg)(5.970x10^{24}kg)}{(6\,378x10^3 m)^2} \right) $
$F_g = 685.66 N$
$F_g = 686 N$|
- b. Calculate the acceleration of the human The acceleration of the human is centripetal, toward Earth's center.
3 sf Fg=m⋅agF_g = m \cdot a_g
m=70.0kgm = 70.0 kg
Fg=685.66NF_g = 685.66 N
r=6 378x103mr = 6\,378x10^3 m		 ag=Fgma_g = \frac{F_g}{m}
ag=685.66N70.0kga_g = \frac{685.66 N}{70.0 kg}
ag=9.795142857...ms2a_g = 9.795142857... \frac{m}{s^2}
ag=9.80ms2a_g = 9.80 \frac{m}{s^2}